Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $k = \dfrac{z + 10}{z^2 + 11z + 10} \times \dfrac{5z^2 - 25z - 30}{-3z - 24} $
First factor out any common factors. $k = \dfrac{z + 10}{z^2 + 11z + 10} \times \dfrac{5(z^2 - 5z - 6)}{-3(z + 8)} $ Then factor the quadratic expressions. $k = \dfrac {z + 10} {(z + 1)(z + 10)} \times \dfrac {5(z + 1)(z - 6)} {-3(z + 8)} $ Then multiply the two numerators and multiply the two denominators. $k = \dfrac {(z + 10) \times 5(z + 1)(z - 6) } { (z + 1)(z + 10) \times -3(z + 8)} $ $k = \dfrac {5(z + 1)(z - 6)(z + 10)} {-3(z + 1)(z + 10)(z + 8)} $ Notice that $(z + 1)$ and $(z + 10)$ appear in both the numerator and denominator so we can cancel them. $k = \dfrac {5\cancel{(z + 1)}(z - 6)(z + 10)} {-3\cancel{(z + 1)}(z + 10)(z + 8)} $ We are dividing by $z + 1$ , so $z + 1 \neq 0$ Therefore, $z \neq -1$ $k = \dfrac {5\cancel{(z + 1)}(z - 6)\cancel{(z + 10)}} {-3\cancel{(z + 1)}\cancel{(z + 10)}(z + 8)} $ We are dividing by $z + 10$ , so $z + 10 \neq 0$ Therefore, $z \neq -10$ $k = \dfrac {5(z - 6)} {-3(z + 8)} $ $ k = \dfrac{-5(z - 6)}{3(z + 8)}; z \neq -1; z \neq -10 $